Problem: The grades on a language midterm at Covington are normally distributed with $\mu = 72$ and $\sigma = 3.5$. Kevin earned a $69$ on the exam. Find the z-score for Kevin's exam grade. Round to two decimal places.
Answer: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Kevin's exam grade by subtracting the mean $(\mu)$ from his grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{69 - {72}}{{3.5}}} $ ${ z \approx -0.86}$ The z-score is $-0.86$. In other words, Kevin's score was $0.86$ standard deviations below the mean.